Contents
PI
PI.1
PI.2
PI.3
PI.4
PI.5
PI.6
PI.7
PI.8
PI.9
PI.10
PI.11
Part One 1
Size of the retinal image 1
Gullstrand Eye model 3
Reflectance of the cornea 4
Radius of curvature 4
Eye length 6
Stereoscopic vision 8
Resolving power of the eye 10
Refracting errors 12
Refractive errors 20
Chromatic aberrations 21
Stereo-camera system 22
Part One
1
PI.1
Size of the retinal image
The retinal image size can be calculated via
(3.7)
|hI | = κ · N0 I00
|hI | = |h0 | ·
N0 I00
O0 N
,
(3.8)
where κ = −hO /O0 N (Section 2.1.3). For an object which is located at
infinity (relaxed eye) and with a mean refractive power of the eye of 60 D,
we may derive the relation
|hI | ≈ 16.667 mm · κ .
(3.9)
If κ is given in degrees, the relation can also be expressed by:
|hI | ≈ 0.291 mm · κ°
(3.10)
with κ = −hO /O0 V (Figure 2.3).
1. Please derive Eq. (3.9) or (3.10).
2. Calculate the size of the retinal image for a tower (50 m high at a distance
of 1 km), a person (1.8 m high at a distance of 10 m), a thumbnail (diameter of 2 cm at a distance of 60 cm), and for the full moon (κ = 0.5°).
Solution:
1. From Eq. (3.7), |hI | = κ · N0 I00 is given. When the object is far away (relaxed
eye/far vision), according to Figure 2.3 we can approximate
N0 I00 ≈ N0 F0 .
(SI.1)
Then, from Eqs. (2.6) and (2.3), we conclude that
N0 F0 = FP =
1
.
2
Deye
(SI.2)
The mean refractive power of a human eye is 60 D, so Deye = 1/f 0 = 60 D,
which means that
f0 =
1
≈ 16.667 mm.
60 m−1
Substituting Eqs. (SI.2) into (SI.1), we find
N0 I00 ≈ f 0 .
(S1.3)
2
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
Accordingly, we have for small angles κ :
|hI | = 16.667 (mm) · κ (rad) .
(3.9)
Converting radians into degrees leads to
1◦ =
π
1◦
π
rad,
=
≈ 0.01745 .
180
1 rad
180
Equation (SI.1) can thus also be written as
|hI | = 16.667 (mm) · (κ0 · 0.01745) ≈ 0.291 mm · κ◦ .
(3.10)
2. Tower: height of 50 m, distance of 1000 m. Hence, we obtain
height
50
arctan
= arctan
≈ 0.05 rad
distance
1000
and thus |hI | = 16.67mm · 0.05 rad ≈ 0.83mm.
Person: height of 1.8 m, distance of 10 m. Hence, we obtain
height
1.8
arctan
≈ 0.178 rad
= arctan
distance
10
and thus |hI | = 16.67 mm · 0.1781 rad ≈ 2.7 mm.
Thumb: diameter of 2cm, distance 60 cm. Hence, we obtain
0.02
diameter
= arctan
≈ 0.033 rad
arctan
distance
0.6
and thus |hI | = 16.67 mm · 0.033 rad ≈ 0.56 mm.
Moon: κ = 0.5◦ , |hI | = 0.291 · κ0 ≈ 0.145 mm. The moon thus appears
roughly as a third in size as compared to the thumb.
Part One
3
PI.2
Gullstrand Eye model
Calculate the position of the eye’s entrance pupil relative to the corneal
vertex. Please do also calculate the diameter of the entrance pupil relative
to the iris aperture by using the Gullstrand Eye model #1.
Solution:
iris
entrance
pupil
optical axis
Figure SI.1 Position of iris and entrance pupil relative to the lens
The entrance pupil is the image of the iris by the cornea. From Figure 2.13, the
following parameters are given:
n0 = 1, s = −3.6 mm, n = 1.336 .
We know from Eq. (2.25) that the total refractive power of the cornea amounts to Dc0
= 43.06 D. With the modified imaging equation (A14), we can write
n0
n
n0
− = 0 = Dc0 ,
0
s
s
f
1
1.336
−
= 43.06 D,
s0
−3.6 mm
which leads to s0 = −3.049 mm. Because of the negative sign, both the object and
the image are “behind” the cornea. The entrance pupil is then located at 3.05 mm
behind the corneal vertex.
The angular magnification (A15) is given by
β=
−3.049 · 1.336
s0 n
=
≈ 1.13 .
sn0
−3.6
Assuming that d0 is the diameter of the entrance pupil and d is the diameter of the
iris, it follows that
β=
d0
→ d0 ⇒ β · d = 1.13 d .
d
4
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
Thus, the diameter of entrance pupil is about 13 % larger than the iris diameter.
PI.3
Reflectance of the cornea
Calculate the reflectance of the cornea at the vertex with the Fresnel equation
(A4).
Solution:
The Fresnel equation (A4) is derived from the more general Fresnel equation (A65)
for the special case of a normal incident ray (γ = 0◦ ) and neglected absorption by
R=
n − n0
n + n0
2
.
For the cornea, we have n = 1 and n0 = 1.376 according to the Exact Gullstrand Eye
# 1 model shown in Figure 2.13. Thus, we have
R=
n − n0
n + n0
2
=
1 − 1.376 2
1 + 1.376
= 0.025.
The reflectance of the cornea at the vertex for a normal incident ray is therefore 2.5%.
PI.4
Radius of curvature
Variations of the corneal radius of curvature ∆rC mean a change of the
corneal refractive power ∆D which can be calculated via
∆D ≈ −
∆rC · Dc0 2
nc − 1
(3.11)
where Dc0 is the refractive power of the corneal front surface and nc the refractive index of the cornea.
a) Please derive Eq. (3.11).
b) Verify the following statement for an emmetropic Gullstand Eye: A variation of the corneal radius of curvature by ±0.1 mm changes the eye’s
refractive power by approximately ±0.6 D.
Part One
5
Solution:
1. In the paraxial approximation according to Eq. (2.22), the refractive power of a
spherical surface is
D0 =
n0 − n
,
rC
(SI.4)
where n and n0 are the refractive indices of the media on the incident and refracted
side, respectively. Taking the derivative of Eq. (SI.4) with respect to rC , we obtain
∆r · (n0 − n)
dD0
n0 − n
= − 2 , or in approximation ∆D0 = − C 2
.
drC
rC
rC
(SI.5)
Re-arranging Eq. (SI.4), yields
1
D0
.
= 0
rC
n −n
(SI.6)
By substituting Eq. (SI.5) into Eq. (SI.6), we obtain
∆D0 = −
∆rC · D02
.
n0 − n
(SI.7)
In the case of the cornea, where D0 = Dc0 (refractive power of the corneal front
surface), n0 = nc (refractive index of cornea), r = rC (curvature of cornea), and
n = 1 (refractive index of air), we finally get
∆D0 ≈ −
∆rC · Dc0 2
.
nc − 1
(3.11)
2. We can calculate the refractive power of the corneal front surface by using Eq.
(2.23) which leads to ∆Dc0 = 48.33 D.
Using the given data ∆rC = ±0.1 mm and nc = 1.376, it follows that
∆D0 ≈ −
(±10−4 m) · (48.83m−1 )2
∆rC · Dc0 2
=−
= ∓0.63 D .
nc − 1
1.376 − 1
Thus, a variation of the corneal radius of curvature by ± 0.1 mm changes the eye’s
refraction (refractive power) by approximately ∓ 0.6 D. An increase of the radius
of curvature by 0.1 mm thus reduces the refractive power by more than a half a
diopter, that is, an emmetropic eye becomes hyperopic.
6
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
PI.5
Eye length
A small variation on an emmetropic eye’s axial length ∆Leye with a refrac0
tive power of Deye
means a change of refraction by ∆D. We may approximate this change with
∆Afar = −∆D0 ≈ −
0 2
∆Leye · Deye
.
n
(3.12)
1. Please derive Eq. (3.12).
2. Verify the following statement for an emmetropic Gullstrand Eye with
0
Deye
= 60 D and n = 1.336: The variation of the eye length by ± 0.37
mm changes the eye’s refraction by approximately ∓ 1 D.
Solution:
1. According to the lens maker’s equation (A14), we have
n0
n
n0
− = 0 .
0
s
s
f
(SI.8)
Applying this formula to the eye’s optical system, we get
n0
0
− Afar = Deye
Leye
(SI.9)
0
in which Deye
is the total refractive power of the eye and Afar = n/s = 1/s the far
point refraction. Leye denotes the eye length for the emmetropic eye. A change in
Leye by a small amount ∆Leye results in a change of the far point refraction (i.e.,
the refractive status of the eye) by ∆Afar . Therefore, the eye’s refractive power is
given by
n0
0
− (Afar + ∆Afar ) = Deye
,
Leye + ∆Leye
from which we can easily obtain
∆Afar =
=
n0
0
− (Afar − Deye
)
Leye + ∆Leye
n0 · ∆Leye
n0
n0
−
=−
,
Leye + ∆Leye
Leye
(Leye + ∆Leye ) · Leye
where we used Eq. (SI.9).
As we assume small changes of the eye length (∆Leye Leye ), we may use the
following approximation:
∆Afar ≈ −
n0 · ∆Leye
.
L2eye
(SI.10)
Part One
7
In the case of an emmetropic eye (Afar = 0), we obtain from Eq. (SI.9)
Leye =
n0
0
Deye
(SI.11)
Inserting Eq. (SI.11) into Eq. (SI.10), we finally obtain
∆Afar ≈ −∆Leye
0 2
Deye
n0
(SI.12)
An increase of the eye length of an emmetropic eye leads to a negative refractive
change, that is, the eye becomes myopic. A decrease of the eye length of an
emmetropic eye leads to a positive refractive change, that is, that the eye becomes
hyperopic.
0
2. Substituting given data Deye
= 60 D, n = 1.336 and ∆Leye = ± 0.37 mm, we find
∆Afar ≈ −
0 2
Deye
∆Leye
(60 D)2 · (±0.37 mm)
=−
≈ ∓0.997 ≈ ∓1
n0
1.336
Thus, a variation of the eye length by ± 0.37 mm changes the eye’s refraction by
approximately ∓ 1 D.
8
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
PI.6
Stereoscopic vision
In order to check the stereoscopic vision of an eye, real and virtual test objects
are used.
1. A real test object shall be used up to a stereo angle of ε = 500 . What
minimum stereoscopic depth perception ∆Lmin must this object have if
it is viewed from a distance of 5 m and the interpupillary distance is PD
= 62 mm.
2. A virtual stereoscopic test object consists of two identical test objects Tl
and Tr (e.g., stripes or triangles). These objects are horizontally arranged
(distance ∆y = 20 mm) above or below a central focus object F (e.g.,
circle). The three objects Tl , Tr and F lie all in one test plane which is
perpendicular to the viewing direction and located at a distance of 5 m.
An optical system ensures that each eye of the patient can only see one test
object. In this regard, we can distinguish between the following cases:
• symmetric allocation, that is, Tl (Tr ) is seen by the left (right) eye, and
• asymmetric allocation, that is, Tl (Tr ) is seen by the right (left) eye.
Due to the small relative shift of both identical images on the retina, the
patient perceives a virtual object T which seems to float behind and in
front of the test plane at a distance ∆L (virtual stereoscopic effect).
a) Is the normal stereoscopic resolution sufficient to have a threedimensional impression?
b) At what distance behind and in front of the test plane does a patient
(with normal stereoscopic vision) see the test objects (interpupillary
distance PD = 65 mm)?
c) Which allocation do we have to choose, if the patient shall perceive a
floating test object located in front of the test plane?
d) Calculate the relative local shift ∆s of the retinal images for an eye
with a refractive power of 60 D.
Solution:
1. We use the geometry of Figure 2.11 with a stereo angle of ε = 500 = 2.424 ×
10−5 rad, the viewing distance L = 5 m and the interpupillary distance PD =
62 mm. According to Eq. (2.18), the minimum stereoscopic depth perception with
a small stereo angle is
∆Lmin =
εL2
.
PD
Inserting the given data into Eq. (2.18) leads to
∆Lmin =
500 · (5 m)2
2.424 × 10−5 rad · (5 m)2
εL2
=
=
= 9.77 mm .
PD
62 mm
62 mm
Part One
9
The test object thus has to provide a depth structure of about 1 cm to become
resolved by the naked eye from a 5 m distance.
2. a) Using again Eq. (2.11) we set ∆y = sp = 20 mm and L = 5 m. Inserting the
given data into Eq. (2.15), we get a stereo angle of
s p
ε = 2 arctan
≈ 0.004 rad ≈ 0.23◦ = 13.80 .
2L
This minimum stereo angle determines the smallest angle that can be resolved
by the eye and still allows stereoscopic perception. Under appropriate conditions, the human eye has a minimum stereo angle of εmin = 1000 . In our
case, the calculated minimum stereo angle of 13.80 is much greater than 1000 .
Therefore, it is possible to have a 3D impression.
b) The test person can see the test objects at ∆Lb behind the test plane and at
∆Lf in front of the test plane. ∆Lb and ∆Lf are given by Eqs. (2.16) and
(2.17) as
L · sp
,
PD + sp
L · sp
∆Lb =
.
PD − sp
∆Lf =
With sp = 20 nm, L = 5 m and PD = 65 mm, we have
L · sp
5 m · 20 mm
=
≈ 1.18 m ,
PD + sp
65 mm + 20 mm
L · sp
5 m · 20 mm
∆Lb =
=
≈ 2.22 m .
PD + sp
65 mm − 20 mm
∆Lf =
Hence, in this case, the test person (with normal stereoscopic vision) sees the
test objects 1.18 m in front of the test plane or 2.22 m behind it.
c) Considering Figure 2.11, the patient shall perceive a floating test object located in front of the test plane with asymmetric allocation.
d) From exercise I.1, we can use |hI | = 16.667 (mm) with κ = ε. It follows that
s p
ε = 2 arctan
≈ 0.004 rad
2L
⇒ |hI | = 16.667 mm · 0.004 rad ≈ 67 µm .
Therefore, the relative lateral shift ∆s = |hI | of the retinal images is about 67
µm for an eye with a refractive power of 60 D.
10
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
a
a
3a
5a
a
a
3a
Figure SI.2 Standard letter E used to determine the refraction of the eye.
PI.7
Resolving power of the eye
1. In order to determine the refraction of an eye, the standard letter “E” is
placed at a distance of 2 ft (= 6096 mm) from the eye. Calculate the size
of the letter for a visual acuity of V = 1.
2. The retinal image resembles the image shown in the Figure 3.19. Calculate the image size on the retina and compare the result to the distance of
the retinal cones (in the fovea).
3. Sometime in the future, we will be visited by aliens from the planet
XIR2050 whose star emits light only in the red and infrared spectral range
and whose atmosphere allows only near-infrared light (wavelengths between 1 and 1.5 µm) to pass through. The eyes of the aliens are adapted
to these conditions and the aliens’ visual acuity is similar to that of our
eyes. How will they fare on Earth? If you were to be selected to travel to
XIR2050, how would you prepare for your visit? What should you expect
to be faced with on this planet?
Solution:
1. The standard letter E has a height of γ = 5a whereby a is the bar width that
characterizes the resolution limit. Using the distance s = 6096 mm and the critical
angle of resolution of α = 10 , one obtains Using the distance s = 6096 mm and
the critical angle of resolution of α = 10 , one obtains
a=α·s
y = 5a = 5sα = 5 · 6096 mm · 10 = 5 · 6096 mm ·
π
= 8.86 mm .
180 · 60
Part One
11
2. Applying the data from Table 2.1 and using the effective focal length of the eye
feff = N0 F0 = 24.385 − 7.331 ≈ 17.1 mm ,
we obtain the size of the retinal image of the letter E
γ 0 = 5feff α = 25 µm .
Accordingly, the scale is a = γ 0 /5 = 5 µm. This is equal to the Airy diameter in
Eq. (A77) given by
dAiry = 1.22 ·
λfeff
= 5.6 µm
diris
for a pupil diameter of approximately diris = 4 mm and a focal length of the eye
of feff = 17 mm. This corresponds to Rayleigh’s criterion of resolution.
In the fovea, the density of cones is maximum with 140,000 mm−2 . This means
that each cone (assuming for simplicity a squared cross section) takes up an area
of
A=
1
= 7.1 × 10−6 mm2 = 7.1 µm2 .
n
This corresponds to the length of an edge or a distance between cones of
∆x = 2.7 µm. Accordingly, about 2 cones are present per length a at the resolution limit, which is quite consistent with the sampling theorem.
3. Assuming the resolution limit to be an angle of 10 or the Airy diameter (A77), as
was done above, it follows that
a = f α = dAiry = 2.44 ·
⇒ α = 2.44 ·
λf
diris
λ
= const.
diris
In order for the critical angle to remain constant while the mean wavelength is
approximately 1.3 µm/0.55 µm = 2.4× larger, the eye pupil of the alien must
be larger by the same factor. Accordingly, the adaptation of the aliens to the
conditions on their planet leads to a pupil of approx. 9.6 mm for normal vision.
Because of the spectral adaptation of their eyes to the atmospheric transmittance
on their planet, the aliens only see in the near-infrared (NIR) spectral range on
Earth. Since the atmosphere of the Earth has a fairly good transmittance in this
range, objects on Earth are illuminated well between 1 µm and 1.5 µm by sunlight
or daylight and the aliens can see well.
In turn, on the planet XIR2050, the atmosphere allows only the light of their star
above 1 µm to pass through. However, the rods and cones of human eyes are not
sensitive in this spectral range. This means that we would actually see nothing.
Travelling to the planet would be useless unless one would use a strong flashlamp
or a NIR-sensitive camera and a head-mounted imaging display for transformation
of the NIR radiation on XIR2050 to the visible spectrum. This is equivalent to
using NIR imaging devices of the type known from nightvision goggles.
12
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
PI.8
Refracting errors
1. In the case of cataracts, it used to be common in the past to simply perforate the turbid lens and sort of remove it in a surgical process. The eye
was rendered aphakic (i.e., left without eye lens). Where is the image of
a far point in an aphakic eye? Would it have been possible to help this
person with spectacles? How strong would these spectacles have to have
been?
2. Is it true that myopic people see small things better? How much of a
difference is there as compared to a person with normal vision?
3. What is your comment regarding the assertion that myopic people can
see sharp images under water; meaning that they do not need a pair of
goggles?
4. In diving schools, it is taught that you see objects under water 33%
bigger and 25% closer. Prove, if this is really true.
Solution:
1. The refractive power of the aphakic eye is exclusively based on the effect of the
cornea, which is given by Dc0 = 43.06 D (Table 2.1). This refractive power acts
right on the vertex of the cornea located at a distance d = 24.385 mm from the
retina in the normal eye (Figure 2.13)1) .
Using n0 = 1.336 (Table 2.1), that is, the refractive index of the aqueous humor
(vitreous), the image of an object placed at infinity is located at a distance
s0 =
n0
= 31.0 mm
Dc0
(SI.13)
from the corneal vertex. As the eye length is 24.4 mm (Table 2.1), the image
plane is located about 6.6 mm behind the retina, which means that an aphakic eye
is extremely hyperopic.
The refractive power in the vertex plane of the cornea to correct for the extracted
lens would have to be
0
Dcorr
=
1.336
n0
=
mm = 54.78 D.
d
24.385
As the corneal power given by Dc0 = 43.06 D, one would have to add 11.7 D, e.g.,
by increasing the corneal curvature, which is beyond practical limits (see Problem
1) Here we assumed that the cornea can be considered as a thin lens with the principal planes at the front
surface. In reality, the meniscus-shaped corneal structure with the distribution of positive and negative refractive power at the front and the rear side, respectively, has a principal plane which is moved
−0.45 mm left to the rear side. This nearly corresponds to the position of the front vertex. Therefore,
the assumption above is quite accurate.
Part One
13
P10.13).
Placing spectacles with a refractive power of Ds0 at a distance Lc = 16 mm in front
of the eye would result in a refractive power of
0
Dcorr
= Dc0 + Ds0 − Lc · Dc0 Ds0 ,
(SI.14)
that is, the sum of the refractive powers of two thin lenses at a distance Lc . Equation (SI.14) is often called Gullstrand’s Equation. The spectacle lens then must
have a refractive power of
Ds0 =
0
54.78 D − 43.06 D
Dcorr
− Dc0
=
≈ 37.7 D .
1 − Lc · Dc0
1 − 0.016 m · 43.06 D
(SI.15)
This means that strongly positive spectacles would have to be used. However,
the refractive power can be reduced somewhat by placing the spectacles closer to
the eye (see table below). However, there are obvious limits for such thick lenses.
The patients would only be able to view objects within a very small angle (“tunnel
vision”).
α (mm)
Ds0 (D)
12
24.3
10
20.6
2. In the case of myopia, an excess of refractive power (by action of cornea and
lens) exists compared to an emmetropic eye with the same axial eye length (socalled refractive myopia). In another situation, we may compare two eyes, one
emmetropic eye with a length of Lemm and one with identical refractive (corneal
and lens) power but with a length of Lmyo > Lemm (so-called axial myopia).
a) First, we compare the image magnification for an uncorrected myopic person
to an emmetropic person (in both myopic situations).
From the lens equations (A14) and (A15), we easily obtain for the image magnification of the myopic eye
βmyo =
s0myo
s
and for the emmetropic eye
βemm =
s0emm
.
s
We assume for both cases the observed object to be located in the far point s
of the myopic eye. Then, by definition we have
s=
1
Afar,myo
14
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
with the far point refraction on the myopic eye Afar,myo . The magnification
then can be written as
|βmyo | =
s0myo
= Afar,myo · s0myo
s
and
|βemm | =
s0emm
= Afar,myo · s0emm .
s
To obtain sharp images on the retina of an object in the far point of the myopic
eye, the following conditions must be fulfilled
s0myo = Lmyo
and
s0emm = Lemm .
For the myopic eye, this condition is fulfilled per definition via
s0myo =
n0
= Lmyo
Dmyo
with Dmyo being the total refractive power of the myopic eye. In contrast, the
emmetropic eye has to accommodate in order to perceive a sharp image of an
object located at s = 1/Afar,myo . This increases the refractive power of the
emmetropic eye by the factor ∆Dacc . After adequate accommodation, we thus
get
s0emm =
n0
= Lemm
Dmyo + ∆Dacc
and finally for the magnifications
|βmyo | = Afar,myo · Lmyo ,
|βemm | = Afar,myo · Lemm .
In the case of so-called refractive myopia, we have Lmyo = Lemm and consequently identical magnifications.
In the case of so-called axial myopia, we have Lmyo > Lemm . Consequently,
the myopic eye will see an object located in the far point of the myopic eye
which appears slightly larger than an object located at the same position seen
by an adequately accommodated emmetropic eye.
b) Before we can compare the image magnifications, we calculate how to correct
the myopic eye. To this end, we use the optical diagram in Figure S.3 for the
corrected eye (and, for simplicity, assume the lenses to be thin).
The far point Qfar of the myopic eye is located at a finite distance sfar =
1/Afar in front of the eye. In order to correct a myopic eye, the spectacles must
Part One
eye 𝒟myo
spectacle 𝒟s‘
15
retina
light path for myopic eye
with corrective spectacles
Qfar
fs
LC
sfar
Figure SI.3 Refractive error correction with spectacles
have a negative refractive power. The resulting optical system comprising the
spectacles and the myopic eye (reduced in its effect to a thin lens in air having
a refractive power of Dmyo ) can be described as follows:
Lc is the always positive distance of the corrective lens (here assumed to be
extremely thin) to the corneal vertex (see also Figure 5.30) and sfar the far
point distance of the eye. According to Figure SI.3 (note that sfar is negative
in the case of myopia!), we can write
fs = sfar + Lc .
With the far point refraction Afar = 1/sfar we derive for the refractive power
of the spectacles
Ds0 (Lc ) =
1
Afar
=
.
fs
1 + Lc · Afar
(SI.16)
For a myopic eye, Afar can also be written as
Afar = Dideal − Dmyo ,
(SI.17)
with Dideal being the refractive power of the ideal eye (emmetropic eye with
the same axial eye length) and Dmyo being the refractive power of the myopic
eye. Note again that for a myopic eye, Afar is negative, since the far point is
situated in front of the eye (sfar < 0) and consequently Dmyo > Dideal .
The resulting refractive power of the corrected myopic eye obtained by the addition of two refractive elements (spectacle lens, myopic eye) being separated
by a distance Lc (see for derivation Section A.1.3) is given by
Dtot = Dmyo + Ds0 (Lc ) − Lc · Dmyo · Ds0 (Lc ) .
Substituting Eq. (SI.16) into Eq. (SI.18) yields
Dtot = Dmyo +
Lc · Dmyo · Afar
Afar
−
.
1 + Lc · Afar
1 + Lc · Afar
(SI.18)
16
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
Using Eq. (SI.17) to substitute Dmyo , we find
Dtot =
Dideal
.
1 + Lc · Afar
(SI.19)
For example, if the far point is at a distance of 0.2 m in front of the myopic
eye and the spectacle lens is placed at 16 mm in front of the corneal vertex, we
find
Ds0 (Lc ) =
Afar
≈ −5.4 D
1 + Lc · Afar
and
Dtot =
Dideal
59D
≈ 63.7 D .
=
1 + Lc · Afar
1 − 0.016 m · 1 D
(SI.20)
c) Next, we want to compare the image magnification for a corrected myopic eye
in comparison to an emmetropic eye. We define the spectacle magnification
as
ms =
retinal image size of myopic eye with correction
.
retinal image size of an emmetropic eye
(SI.21)
Essentially, we compare two sharp retinal images. For a corrected myopic eye,
the retinal image size of a distant object with an angular size κ is given by
h0myo,corr =
κ
.
Dtot
The corresponding retinal image size of an emmetropic eye is given by
h0emm =
κ
.
Dideal
Thus, Eq. (SI.21) becomes
ms =
=
retinal image size of myopic eye with correction
retinal image size of an emmetropic eye
h0myo,corr
D
= ideal .
h0emm
Dtot
With Eqs. (SI.17) and (SI.19), we finally obtain
ms =
1
.
1 + Lc · Afar
(SI.22)
For a spectacle position of Lc = 16 mm in front of the cornea, we calculate the
following values for various myopic refractions:
Part One
Afar
Ds0
ms
-1
-2
-3
-5
-10
-1.0
-2.1
-3.2
-5.4
-11.9
0.98
0.96
0.95
0.92
0.84
17
Evidently, a corrected myopic person indeed sees objects smaller when viewing through the correcting spectacles in comparison to an emmetropic person.
d) Finally, we want to compare the image magnification for a corrected myopic
eye with an uncorrected myopic eye. For this purpose, we define the spectacle
magnification using the symbols in Figure SI.4 as
retinal image size of myopic eye with correction
retinal image size of a myopic eye without correction
angular size with correction
=
angular site without correction
m0s =
=
αcorr
.
α
(SI.23)
The uncorrected myopic eye views an object with height y at a distance x from
the entrance pupil. The object appears under an angle α = y/x.
α
y
entrance pupil
x
Figure SI.4a Image formation for an uncorrected eye
𝒟s‘
αcorr
y
γ
s
Le
γ
y
entrance pupil
s‘
Figure SI.4b Image formation by a correcting lens
Now, we put a lens with a refractive power of Ds0 in front of the eye. The object
should be at a distance s from the lens. The object now subtends an angle γ .
18
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
From Figure SI.4b, we can deduce
y0 = y ·
s0
s
and
αcorr =
s0
y0
− Le
in which we can substitute y 0 and where Le is the distance of the lens from the
entrance pupil plane of the myopic eye. With Eq. (SI.23), we find
m0s =
αcorr
s0 x
sx
=
=
,
0
α
s(s − Le )
1 − Le S 0
in which we used S = s−1 and S 0 = s0−1 . Further, we can write the image
equation (A14) for the situation of Figure SI.4 as
S 0 = S + Ds .
If we now assume a distant object, for which x → ∞, S → 0, x·S → 1, S 0 →
Ds0 , we finally obtain
m0s =
1
.
1 − Le · Ds
Using Dideal = 59 D and Le = Lc + 3mm = 19 mm, we find the values
shown in the following Table for the spectacle magnification as a function of
the degree of short-sightedness Afar :
Afar
Ds0
m0s
-1
-2
-3
-5
-10
-1.0
-2.1
-3.2
-5.4
-11.9
0.98
0.96
0.94
0.91
0.82
Evidently, a myopic eye indeed sees objects correspondingly smaller when
viewing through the correcting spectacles (but at least sharply). Moreover,
we recognize that only in the case of a mild refraction error, the power of the
spectacles is approximately equal to the refractive error.
3. Assuming, (for approximation purposes,) the refractive index of water to be n =
1.334, the refractive effect of the front surface of the cornea (Da0 = 48.8 D; see
Eq. (2.23)) is almost not effective under water, since there is only a small difference
in refractive index at the front of the eye.
Part One
19
Using Eq. (2.23) and data from the Exact Gullstrand Eye # 1 (Table 2.1), we get
for the refractive power of the corneal front surface in water
Da0 =
1.375 − 1.334
= 5.45 D .
0.0077
According to Eq. (2.24), we get for the refractive power of the corneal back surface
Dp0 =
1.336 − 1.376
= −5.88 D .
0.0068
With Eq. (2.20), the total refractive power of the cornea in water follows as
0
Dc,in
water = 5.45 D + (−5.88) D −
0.0005
(5.45 · (−5.88) D
1.336
= −0.42 D ≈ 0 D .
Thus, only the refractive power Dl = 19.1 D of the lens (relaxed) or Dl = 33.1 D
(accommodated) remains in this case.
According to Table 2.1 (Exact Gullstrand Eye #1), an emmetropic eye has a total
refractive power of about 59 D in air (relaxed vision). In water, the power is
reduced by about 40 D ≈ 59 D − 19 D and if compensated by maximum accommodation still by about 26 D ≈ 59 D − 33 D. As the eye length is constant, this
power reduction leads to significant (under-water) hyperopia. The same is true for
a myopic eye. However, the (under-water) hyperopia is reduced by the refraction
(refractive error) in air, if the refractive error originates from axial myopia (i.e., a
longer eye length). In addition, in order to see a noticeable (under-water) hyperopia reduction effect, the (in-air) myopia must be extremely high (above −20 D).
There is no difference at all between a myopic and an emmetropic person’s vision
under water if the myopia in air is caused by a smaller radius of curvature of the
cornea.
4. As we know from Eq. (A15), the magnification is given by
β=
ns0
.
n0 s
In this case, s0 , n0 and s are constant. The magnification in air can thus be written
as
βair =
and under water as
βw =
s0
n0 s
1.33s0
.
n0 s
Thus, when s (the distance between the object and the people’s eyes) is constant,
people see objects under water in comparison to air as
βw
= 1.33 ,
βair
20
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
that is 33% bigger.
When people see an object showing the same apparent size under water as in
air, the magnification is
βair = βw .
Accordingly,
nw s0
nair s0
= 0
.
0
n sair
n sw
Due to the structure of eyes, s0 and n0 are always constant. Consequently, nair /sair
should be equal to nw /sw .
Substituting nair = 1 and nw = 1.33, we obtain
sw = 1.33 · sair or sair = 0.75 · sw .
Thus, an object appearing at the same size under water and in air, seems to be
25% closer in water than in air.
PI.9
Refractive errors
1. Use the Gullstrand Eye model to calculate the power of spectacles needed
to correct an eye if the far point is located 45 cm in front of the eye.
Because of the frame of spectacles, the glasses are placed at a distance of
15 mm in front of the corneal vertex.
2. The same spectacle glasses as in 1) have inadvertently been mounted in
a frame so that we now have a distance from the vertex of only 10 mm.
Does this improve or worsen the correction of the refractive error?
3. Draw a conclusion from 2) regarding how a contact lens would have to
be designed.
Solution:
1. The far point sfar is located 45 cm in front of the eye. This corresponds to a far
point refraction (refractive error) of
Afar =
1
1
=
≈ −2.22 D .
sfar
−45cm
According to Problem PI.8, the required back vertex power of the corrective spectacle glass is given by
Ds0 (Lc ) =
Afar
.
1 + Lc · Afar
Part One
21
As the spectacles are placed at a vertex distance of Lc = 15 mm from the cornea,
we have
Ds0 (Lc ) =
−2.22
= −2.3 D .
1 + 0.015(−2.22)
2. If the distance of the spectacles from the corneal vertex is only Lc = 10 mm, the
correction results as
Ds0 (Lc ) =
−2.22
= −2.27 D .
1 + 0.01(−2.22)
Hence, the correction is slightly poorer. However, this will hardly be noticeable.
The sign is disadvantageous, though, since the additional error cannot be corrected by means of accommodation.
3. For a contact lens of negligible thickness, we have Lc ≈ 0. Thus, the required
back vertex power of the contact lens is almost equal to the refractive error ( ≈
−2.22 D) of the eye.
PI.10
Chromatic aberrations
The eye shows some notable chromatic aberration of almost 2 D in the visible spectral range. Why do we generally not notice this, whereas an optical instrument (e.g. photo camera) with similar chromatic aberration would
be unusable? Could the different width of the blue-white-red stripes of the
French flag (Tricolore) have anything to do with this?
Solution:
An axial chromatic error of 2 D between red and blue is clearly above the bothersome
limit of a refraction error (the limit is at about 0.25 D). However, the corresponding
color receptors (L, S, M cones) are rather elongated in shape and somewhat staggered
in their depth. Moreover, these types of blur are usually corrected by the brain based
on experience. Therefore, the effect of longitudinal chromatic aberration in the eye
is less bothersome than in technical systems for imaging.
The difference in axial depth of the colored images of 2 D causes a difference in focal
intercept (Figure 6.11) of
∆s = sred − sblue =
1
1
m−
m = 0.55 mm .
59.5
61.5
Theoretically, this defocussing causes a chromatic magnification difference (lateral
chromatic aberration, see Section A.1.9) of
∆y 0
∆s
=
= 60.5 mm−1 · 0.55 mm = 3.3% .
y0
f
22
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
This magnitude is already noticeable by eye. However, in reality, the circumstances
are far more complicated:
1. The principal nodal points vary by color. The estimation presented above is thus
simplified.
2. The retina is a curved “receiver” which has an impact on the geometric conditions.
3. Usually, extensive image fields are viewed with a rotating eye in order to optimally
utilize the focusing range of the fovea. Accordingly, the field angle produced in
the eye is not very large, and the difference in chromatic magnification is relatively
small.
4. For finite field angles, we have rod receptors in the peripheral range. These are
not color-sensitive.
Accordingly, the apparent difference in the width of the stripes of the Tricolore is
most likely not only a consequence of the lateral chromatic aberration of the eye.
PI.11
Stereo-camera system
Let us consider the special case of a stereo-camera system. Both cameras
have equal parameters (in particular equal focal lengths f ), the image planes
of the two cameras are co-planar, and the x axes of the image planes are
parallel to the baseline (Figure 3.20). The distance between the two cameras
is b = 200 cm. The cameras use 1/200 CCD chips with 1024 horizontal
pixels. A camera system system of this type shall be used to track objects
(e.g. surgical instruments) in a volume (a3 = 50 x 50 x 50 cm3 ) around a
patient’s head from a distance z ≈ 3 m.
1. Calculate the stereo-disparity (difference between xL and xR ) for various
objective focal lengths f .
2. Which objective focal length would you recommend to achieve a tracking
volume of maximum size or maximum point accuracy?
3. Calculate the maximum attainable (lateral) resolution of the system at the
optimal objective focal length. What causes the resolution to be less in
reality? How can the resolution be increased?
4. Compare this resolution to that of an acoustical tracking system with a
frequency of 50 kHz. Which phase measuring accuracy must be at least
attained with an ultrasound system?
Solution:
In Figure 3.20, f is the focal length, x the lateral object coordinate. In this case, we
have the middle of the camera axes as the origin.
Part One
23
a
tracking
volume
x
z
b/2
plane of
camera pupils
f
xL
xR
plane of
CCD chips
x
b
Figure 3.20 Geometry of a stereo-camera system discussed in Problem I.??.
1. According to the theorem of Thales, the following equations apply to similar triangles:
x − b/2
xL
=
,
f
z
x + b/2
xR
=
.
f
z
Accordingly, the mean of the stereodisparity of the center of the volume is
∆x = xR − xL =
f ·b
.
z
(SI.25)
Using b = 200 cm and z = 300 cm, the following examples can be calculated:
1)
2)
3)
f = 10 mm = 1 cm
f = 20 mm = 2 cm
f = 30 mm = 3 cm
⇒ ∆x = 0.67 cm
⇒ ∆x = 1.33 cm
⇒ ∆x = 3.3 cm
2. Considering just one camera (i.e., the camera on the right), we obtain – similar to
the preceding exercise – for the outer point of the measuring volume at a distance
x = a/2:
xR,max
b/2 + a/2
=
f
z
Solutions to Problems – Optical Devices in Ophthalmology and Optometry
24
from which follows that
2xR,max · z
−b .
(SI.26)
f
The maximum permissible camera coordinate xR,max is determined by the size
a=
of the sensor.
According to Eq. (SI.26), a increases with decreasing f . As a consequence, a
large transverse measuring volume can be obtained with small focal lengths.
Conversely, according to Eq. (SI.25), a large focal length results in the transverse
disparity being large, which results in high accuracy.
3. The optimal focal length is the largest focal length that still permits the measuring
range of the parameter a. This mean that, for x = a/2, the transverse disparity is
equal to half the chip width xR,max = D, that is, xR,max = D/2. Accordingly,
the optical focal length is given by Eq. (SI.26) as
f=
2xR,max · z
z·D
=
= 15.2 mm ,
a+b
a+b
with z = 300 cm, b = 200 cm, a = 50 cm and D =
From
1 00
2
(SI.27)
= 1.27 cm.
x + 2b
xR
=
f
z
b
z · xR
−
⇒x=
f
2
follows by differentiation that
dx
z
.
=
dxR
f
D
Using a pixel size of p = 1024
= 12.4 µm as the accuracy of the camera ∆xR and
the optical focal length f from Eq. (SI.25), the total accuracy results as
∆x =
z
= 2.45 mm .
f · ∆xR
(SI.28)
In reality, the accuracy is lesser due to
•
•
•
•
a residual camera error,
a non-symmetrically positioned volume,
a depth variation of the volume, and
inaccuracies due to tracking markers being finite.
The accuracy can be increased by
• a reduced size of the volume,
• a larger chip and simultaneously larger focal length, and by
• the detection of markers at subpixel level by means of image processing.
4. An ultrasound system with 50 kHz has a wavelength of 6.9 mm in air. It can thus
easily compete in terms of accuracy with an optical system.